I have a limaçon with the polar equation $r=a+b\cos(\theta)$. I want to strecth it in the direction of the vector $W=(c,d)$ and squeeze it in the opposite direction of $W$. I have a figure here which may help. The black one is the origin limaçon and the green one is the squeezed and sctetched one. Here $a=3.2$ and $b=2$.
There is a solution for the case that I squeeze or scretch in the direction of $W$.
Now, the question is:
what would be the equation of this new closed curve? will this new closed curve still will be a limaçon?
I will appreciate any comment or solutions.
tl;dr: No, a stretched limaçon is not a limaçon.
Let's be a bit more "generous" and call a limaçon any polar graph $$ r = a + b\cos(t + t_{0}), $$ i.e., that can be obtained by rotating the polar graph $r = a + b\cos t$ about the origin by $t_{0}$. A general limaçon therefore has parametric form \begin{align*} x(t) &= (a + b\cos(t + t_{0}))\cos t, \\ y(t) &= (a + b\cos(t + t_{0}))\sin t. \end{align*}
It appears stretching amounts to applying a specific "orthogonally diagonalizable" linear transformation. Again, let's be more flexible and assume we're allowed to apply an arbitrary invertible linear transformation $$ \left[\begin{array}{@{}c@{}} u \\ v \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} A & B \\ C & D \\ \end{array}\right] \left[\begin{array}{@{}c@{}} x \\ y \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} Ax + By \\ Cx + Dy \\ \end{array}\right],\quad AD - BC \neq 0. $$ The parametric equations for a "stretched" limaçon are therefore \begin{align*} u(t) &= (a + b\cos t)(A\cos t + B\sin t), \\ v(t) &= (a + b\cos t)(C\cos t + D\sin t). \end{align*} This is not a "limaçon" in the stated sense unless the linear transformation is a Euclidean motion (rotation about the origin or reflection about a line through the origin) followed by scaling about the origin: We must have \begin{align*} A\cos t + B\sin t &= E\cos(t + t'), \\ C\cos t + D\sin t &= E\sin(t + t') \end{align*} for some real $E$ and $t'$. The remaining details are a little tedious to write out, but follow from addition formulas for the trig functions, which lead to $A^{2} + B^{2} = E^{2} = C^{2} + D^{2}$ and $AC + BD = 0$, so that the transformation matrix has orthogonal columns of equal length.
As for the equations of the transformed curve, multiplying the equation of the limaçon by $r$ gives $$ x^{2} + y^{2} = r^{2} = ar + br\cos t = ar + bx. $$ Rearranging and squaring, $$ (x^{2} + y^{2} - bx)^{2} = (ar)^{2} = a^{2}(x^{2} + y^{2}). $$ This quartic equation in $(x, y)$ may be expressed in terms of $u$ and $v$ by noting that \begin{align*} \left[\begin{array}{@{}c@{}} x \\ y \\ \end{array}\right] &= \left[\begin{array}{@{}cc@{}} A & B \\ C & D \\ \end{array}\right]^{-1} \left[\begin{array}{@{}c@{}} u \\ v \\ \end{array}\right] \\ &= \frac{1}{AD - BC} \left[\begin{array}{@{}rr@{}} D & -B \\ -C & A \\ \end{array}\right] \left[\begin{array}{@{}c@{}} u \\ v \\ \end{array}\right] \\ &= \frac{1}{AD - BC} \left[\begin{array}{@{}c@{}} Du - Bv \\ Av - Cu \\ \end{array}\right]. \end{align*} The details are again a bit tedious, but the result is a quartic polynomial, and the zero locus has the same qualitative shape (e.g., a cusp or crossing at the origin becomes a cusp or crossing at the origin) even if it is not strictly a limaçon.